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3 years ago

What is the temperature of 4.5 moles of a gas that occupies 50. mL at 1.35 atm?

Chemistry

1 answer:

devlian [24]3 years ago

7 0

Answer:

$T = 0.182$ Kelvin

Explanation:

As we know that

$PV = nRT\\$

Where P is the pressure in atmospheric pressure

T is the temperature in Kelvin

R is the gas constant

V is the volume in liters

$R = 0.08206$

Substituting the given values in above equation, we get -

$1.35 * \frac{50}{1000}= 4.5 * 0.08206* T\\$

On rearranging, we get

$T = \frac{1.35*50}{1000*0.08206*4.5} \\$

$T = 0.182$ Kelvin

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Which of the following best describes the process of electroplating?

sweet-ann [11.9K]

I think the correct answer would be B. The process of electroplating uses a direct current in order to give a cheap metal the apperance and resistance of a more expensive metal by coating it with a certain metal. The current reduces the dissolved metal ions to form a thin coating on a surface.

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3 years ago

g Ammonia has been studied as an alternative "clean" fuel for internal combustion engines, since its reaction with oxygen produc

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$\text{Ammonia has been studied as an alternative "clean" fuel for internal combustion}$

$\text{engines, since its reaction with oxygen produces only nitrogen and water vapor,}$

$\text{and in the liquid form it is easily transported. An industrial chemist studying this}$

$\text{reaction fills a} \ \mathbf{100 \ L }\ \text{tank with} \ \mathbf{8.6 \ mol} \ \text{of ammonia gas and} \ \mathbf{28 \ mol} \ \ \text{of oxygen gas, }$

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$\text{ammonia at the final temperature of the mixture. Round your answer to 2 significant digits.}$

Answer:

Explanation:

From the correct question above:

The reaction can be represented as:

$\mathbf{4 NH_3_{(g)}+ 3O_{2(g)} \iff 2N_{2(g)}+ 6H_2O_{(g)} }$

From the above reaction; the ICE table can be represented as:

$\mathbf{4 NH_3_{(g)}+ 3O_{2(g)} \iff 2N_{2(g)}+ 6H_2O_{(g)} }$

I (mol/L) 0.086 0.28 0 0

C -4x -3x +2x +6x

E 0.086 - 4x 0.28 - 3x +2x +6x

At equilibrium;

The water vapor = $\dfrac{2.6 \ mol}{100 \ L} = 6x$

$x = \dfrac{2.6}{100} \times \dfrac{1}{6}$

$x = 0.00433$

$\text{equilibrium constant} ({k_c}) = \dfrac{ [N_2]^2 [H_2O]^6 }{ [[NH_3]^4] [O_2]^3 }$

$\implies \dfrac{(2x)^2 (6x)^6}{(0.086-4x)^4\times (0.28-3x)^3} \\ \\$

Replacing the value of x, we have:

$K_c = \dfrac{4 \times 46,656 \times x^8}{(0.086-4x)^4\times (0.28 -3x)^3} \\ \\ K_c = \dfrac{4 \times 46656 \times (0.00433)^8}{(0.06868)^4(0.26701)^3} \\ \\ K_c = \mathbf{5.4446 \times 10^{-8}}$

$K_c = \mathbf{5.5 \times 10^{-8} \ to \ 2 \ significant \ figures}$

5 0

3 years ago

By titration, it is found that 31.7 mL of 0.145 M NaOH(aq) is needed to neutralize 25.0 mL of HCl(aq). Calculate the concentrati

arlik [135]

Answer:

0.184 M

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

HCl + NaOH —> NaCl + H2O

From the balanced equation above, the following data were obtained:

Mole ratio of the acid, HCl (nA) = 1

Mole ratio of the base, NaOH (nB) = 1

Next, the data obtained from the question. This includes:

Volume of the base, NaOH (Vb) = 31.7 mL Molarity of the base, NaOH (Mb) = 0.145 M

Volume of the acid, HCl (Va) = 25.0 mL

Molarity of the acid, HCl (Ma) =?

Finally, we shall determine the molarity of the acid (HCl) as shown below:

MaVa /MbVb = nA/nB

Ma × 25 / 0.145 × 31.7 = 1

Cross multiply

Ma × 25 = 0.145 × 31.7

Ma × 25 = 4.5965

Divide both side by 25

Ma = 4.5965 / 25

Ma = 0.184 M

Therefore, the molarity of the acid (HCl) is 0.184 M

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Answer:

4) pink phenolphthalein

Explanation:

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3 years ago